   I have found at http://mathworld.wolfram.com/Maclaurin-CauchyTheorem.html the

following claim:

Theorem (Maclaurin-Cauchy). If f(x) is positive and decreases to 0, then an Euler constant

γ_{f} ≡ lim_{n→∞} [∑_{k=1}^{n} f(k) − ∫_{1}^{n} f(x)dx]

can be defined.

For example, if f(x) = 1/x, then γ_{f} = γ is just the usual Euler-Mascheroni constant.

I wanted to use this theorem but I didn’t find any other reference except from these websites. Therefore I tried to prove it myself and I figured out that the proof is easy. (Moral of the story: Before looking for books and papers containing the theorem, I should have thought about a way to proof it.)

Proof. We estimate difference between the area under the graph of function f (corresponding to the integral) and the “steps” corresponding to the sum. Let us denote

a_{n} = ∑_{k=1}^{n} f(k) − ∫_{1}^{n+1} f(x)dx.

We see that 0 ≤ a_{n} −a_{n−1} = f(n)− ∫_{n}^{n+1}f(x)dx ≤ f(n)− ∫_{n}^{n+1}f(n+1)dx = f(n)−f(n+1).

The sequence a_{n} is n-th partial sum of a series of positive numbers f(n) − ∫_{n}^{n+1}f(x)dx.

Therefore a_{n} is a non-negative increasing sequence. Moreover it is bounded, since a_{n} ≤ f(1) − f(2) + f(2) − f(3) + … − f(n) = f(1) − f(n) ≤ f(1).

Finally, I have found a proof based on the same idea (although it is written more elegant) in [?, p.194–195]. (Moral of the story: Websearch is as good for you as you’re good in websearch.)

In some situations the above limit can be estimated using Euler summation formula in the following form

∑_{k=0}^{n} f(k) = ∫_{0}^{n} f(t)dt + \frac{f(0) + f(n)}{2} + ∫_{0}^{n}(t − [t] − \frac{1}{2})f′(t)dt

was possible as well, since the function f had continuous first derivative.

References

 [1] J. Franke: Vorlesungsskript Analysis I, www.math.uni-bonn.de/people/franke/A1-4.ps